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# Calculating the equilibrium constant

Calculating the equilibrium constant. I have had trouble with this problem.

The EXACT question asks: Calculate the equilibrium constant at 298K for the following equation.

Sn(+2)(aq) + MnO4(-1)(aq) ===> Sn(+4)(aq) + Mn(+2)

I understand Kc = (Products^(stoic. coefficients))/(Reactants^(stoic. coefficients))
Stoic. coefficients = 1 (1:1)
This is a no brainer, but how do I get a NUMERICAL Kc value?

Perhaps I have to use thermodynamics (which I cannot find deltaG or deltaH for this equation), or perhaps Nernst equation using reduction potentials:

MnO4(-1)(aq) + 8H(+)(aq) +5e ==> Mn(2+)(aq) + 4H2O 1.51v
Sn(+4)(aq) + 2e ==> Sn(+2)(aq) 0.15v
-------------------------------------------------------------------------------------------
Balanced Eq:

5Sn(+2) + 2MnO4(-1)+16H+ ==> 5Sn(+4)+2Mn(+2)+8H2O
--------------------------------------------------------------------------------------------

Or do I use an inverse : K(reverse) = 1/(K(forward)) but it is not a cell.
Perhaps the rule of multiple equilibrium(adding reaction steps)
I need a NUMERICAL answer, NOT just the Kc expression.

If anyone can help me I would appreciate it immensely!

#### Solution Preview

Hi, you need to have some confidence in what you know on these.

You know at equilibrium the cell emf is going to be zero so Eo is going to be 0.509/n log K
It really does not matter if ...

#### Solution Summary

This solution is provided in 156 words. It uses equilibrium and Nernst relations to determine the equilibrium constant.

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